A body accelerates uniformly from rest at 2ms-2. Calculate its velocity when it has traveled a distance of 9m
To calculate the velocity of a body when it has traveled a certain distance under uniform acceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity (which is 0 m/s since the body starts from rest)
a = acceleration
s = distance
Given that the initial velocity (u) is 0 m/s, the acceleration (a) is 2 m/s^2, and the distance (s) is 9 m, we can solve for the final velocity (v).
v^2 = 0^2 + 2 * 2 * 9
v^2 = 0 + 36
v = √36
v = 6 m/s
Therefore, the velocity of the body when it has traveled a distance of 9 m is 6 m/s.
v = √(2as) = √(2*2*9) = 6
To calculate the velocity of a body accelerating uniformly, we can use the following equation:
v = u + at
Where:
v = final velocity
u = initial velocity (in this case, it is 0 since the body starts from rest)
a = acceleration
t = time
Since we want to find the velocity when the body has traveled a distance of 9m, we need to find the time it takes for the body to reach that distance.
To find the time, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
t = time
a = acceleration
In this case, we know that initial velocity u = 0, acceleration a = 2 ms^(-2), and distance traveled s = 9m. Plugging in these values, we get:
9 = (1/2)(2)t^2
Rearranging the equation, we get:
t^2 = (9 * 2) / 1
t^2 = 18
t = √18
Therefore, the time taken to travel a distance of 9m is t = √18 seconds.
Now that we have the time, we can calculate the velocity using the first equation:
v = u + at
v = 0 + 2 * √18 ms^(-1)
Therefore, the velocity of the body when it has traveled a distance of 9m is v = 2√18 ms^(-1).